question_answer

Solve the following differential equation. \[\frac{dy}{dx}=1-x+y-xy\]

Answer:

Given differential equation is \[\frac{dy}{dx}=1-x+y-xy\] which can be rewritten as \[\frac{dy}{dx}=1-x+y(1-x)\] \[\Rightarrow \] \[\frac{dy}{dx}=(1-x)(1+y)\] Now, separating the variables, we get             \[\frac{dy}{1+y}=(1-x)\,dx\] On integrating both sides, we get \[\int{\frac{dy}{1+y}}=\int{(1-x)\,dx}\] \[\Rightarrow \] \[\log |1+y|\,\,=x-\frac{{{x}^{2}}}{2}+C,\] which is the required solution of differential equation.